Suppose 64% of a remote mountain village can taste phenylthiocarbamide PTC) and must therefore have at least one copy of the dominant PTC taster allele. If this population conforms to Hardy- Weinberg expectations for this gene, what percentage of the population must be heterozygous for this trait?

Question

Quizplus · Accepted Answer

48%In a population that conforms to Hardy-Weinberg equilibrium, the frequencies of alleles and genotypes remain constant from generation to generation if no other evolutionary influences are at work. The frequency of the dominant allele (A) plus the frequency of the recessive allele (a) equals 1. If p represents the frequency of the dominant allele and q represents the frequency of the recessive allele, then p + q = 1.Given that 64% of the population can taste PTC and must have at least one copy of the dominant allele, we can infer that 64% represents the combined frequency of the homozygous dominant (AA) and heterozygous (Aa) genotypes. Therefore, p^2 + 2pq = 0.64, where p^2 represents the frequency of the homozygous dominant genotype, and 2pq represents the frequency of the heterozygous genotype.Since q = 1 - p, and we know that 36% of the population cannot taste PTC (and are therefore homozygous recessive, aa), q^2 = 0.36. Taking the square root of both sides gives q = 0.6. Substituting back, p = 1 - q = 1 - 0.6 = 0.4.The frequency of the heterozygous genotype (2pq) is therefore 2(0.4)(0.6) = 0.48, or 48%.

Suppose 64% of a Remote Mountain Village Can Taste Phenylthiocarbamide